What mass of precipitate (in g) is formed when 80.2 mL of 0.500 M AlBr₃ reacts with excess AgNO₃ in the following chemical reaction? AlBr₃(aq) + 3 AgNO₃(aq) → 3 AgBr(s) + Al(NO₃)₃(aq)

Question

What mass of precipitate (in g) is formed when 80.2 mL of 0.500 M AlBr₃ reacts with excess AgNO₃ in the following chemical reaction? AlBr₃(aq) + 3 AgNO₃(aq) → 3 AgBr(s) + Al(NO₃)₃(aq)

Answer 1

AlBr₃(aq) + 3 AgNO₃(aq) → 3 AgBr(s) + Al(NO₃)₃(aq) ... balanced equation

Since there is excess AgNO₃, no need to worry about limiting reactant, as AlBr3 will determine amount of product that can form

moles AlBr₃ = 80.2 ml x 1 L / 1000 ml x 0.500 mol / L = 0.0401 mols

moles AgBr(s) that can form = 0.0401 mol AlBr₃ x 3 mol AgBr / 1 mol AlBr₃ = 0.1203 mols

mass AgBr(s) formed = 0.1203 mols AgBr x 188 g AgBr / mol = 22.6 g AgBr(s) formed (3 sig. figs.)